Karnataka Board PUCPUC Science 2nd PUC Class 12

The Decay Constant of 197 80 Hg (Electron Capture to 197 79 Au) is 1.8 × 10−4 S−1. (A) What is the Half-life? (B) What is the Average-life? (C) How Much Time Will It Take to Convert 25% - Physics

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Sum

The decay constant of `""_80^197`Hg (electron capture to `""_79^197`Au) is 1.8 × 10−4 S−1. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?

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Solution

Given :-

Decay Constant of `""_80^197"Hg" , lambda = 1.8 xx 10^-4 "s"^-1`

(a)

Half-life, `T_"1/2" = 0.693/lambda`

`⇒ T_"1/2" = 0.693/(1.8 xx 10^-4)`

= 3850 s=64 minutes


(b) 

Average life, `T_(av) = T_"1/2"/0.693`

`= 64/0.693`

= 92 minutes


(c)

Number of active nuclei of mercury at t = 0 = N0 = 100

Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N =  75

Now , `N/N_0 = e^(-lambda t)`

Here,

N = Number of inactive nuclei
`N_0` = Number of nuclei at t = 0
`lambda =` Disintegration constant

On substituting the values, we get

`75/100 = e^(-lambdat)`

`⇒ 0.75 = e^(-lambda x)`

`⇒ "In"  0.75 = - lambda t`

`⇒ t = ("In" 0.75)/-0.00018`

= 1600 s

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Chapter 24: The Nucleus - Exercises [Page 442]

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HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 24 The Nucleus
Exercises | Q 19 | Page 442

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